Explanation

(d)
$f(x) = \left| \begin{matrix}
x^{3} & x^{4} & 3x^{2} \\
1 & - 6 & 4 \\
p & p^{2} & p^{3}
\end{matrix} \right|$
$\Rightarrow \frac{d}{dx}f(x) = \left| \begin{matrix}
3x^{2} & 4x^{3} & 6x \\
1 & - 6 & 4 \\
p & p^{2} & p^{3}
\end{matrix} \right|$
$\Rightarrow \frac{d^{2}}{dx^{2}}f(x) = \left| \begin{matrix}
6x & 12x^{2} & 6 \\
1 & - 6 & 4 \\
p & p^{2} & p^{3}
\end{matrix} \right|$
$\Rightarrow \frac{d^{3}}{dx^{3}}f(x) = \ \left| \begin{matrix}
6 & 24x & 0 \\
1 & - 6 & 4 \\
p & p^{2} & p^{3}
\end{matrix} \right|$
$\Rightarrow \frac{d^{4}}{dx^{4}}f(x) = \left| \begin{matrix}
0 & 24 & 0 \\
1 & - 6 & 4 \\
p & p^{2} & p^{3}
\end{matrix} \right|$
$= - 24\ \left| \begin{matrix}
1 & 4 \\
p & p^{3}
\end{matrix} \right| = - 24(p^{3} - 4p)$
Hence, $\frac{d^{4}}{dx^{4}}f(x)$ is a constant.